All categories

3 years ago

How to get the formulas for these compounds

Chemistry

1 answer:

Travka [436]3 years ago

5 0

So what you’re going to do is basically the + and - in each top hand corner is the charge of compound, so for example Li has a charge of +1 while Br has a charge of -1 , to write the formula you need to get the charges to cancel out ( equal zero) so luckily this was easy because -1 +1 =0 ! So it would be LiBr. Though for another example Al has a charge of 3+ while br has a charge of -1 and these do not equal zero, so as a result you have to add more br making the Formula AlBr3! Hope this helps!

You might be interested in

Consider the following specific heats: copper, 0.384 J/g· °C; lead, 0.159 J/g· °C; water, 4.18 J/g· °C; glass, 0.502 J/g· °C. Wh

Reil [10]

Answer:

3. water

Explanation:

Specific Heat: Specific heat of a substance is the energy required to increase temperature of 1 gram of substance by 1°C

Here water has the highest specific heat in the list,so it emit largest amount of heat with lose of small amount of temperature.

So,water can warmed body until its temperature becomes below the body temperature.

5 0

3 years ago

A compound contains potassium, nitrogen, and oxygen. the experimental analysis gave values of 45.942 % potassium and 16.458 % ni

9966 [12]

It should be KNO3 to the best of my understanding

7 0

3 years ago

Read 2 more answers

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <---> N2(g) 3H2(g) At equilibrium, it was fou

Softa [21]

Answer:

Kc for this equilibrium is 2.30*10⁻⁶

Explanation:

Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.

Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is defined as:

$Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }$

In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.

In this case, being:

2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)

the equilibrium constant Kc is:

$Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }$

Being:

[N₂]= 0.0551 M
[H₂]= 0.0183 M
[NH₃]= 0.383 M

and replacing:

$Kc=\frac{0.0551*0.0183^{3} }{0.383^{2} }$

you get:

Kc= 2.30*10⁻⁶

<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>

8 0

4 years ago

The liquid state of matter can be described as

wariber [46]

Having a definite volume but not a definite shape.

3 0

3 years ago

Read 2 more answers

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

andrew-mc [135]

Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options $XeF_{2}$ is the correct examples of linear molecules for five electron groups.

8 0

3 years ago